Edexcel GCE Chemistry 6246/02. June Results Mark Scheme. Chemistry 6246/02. Edexcel GCE

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Edexcel GCE Chemistry 6246/02 June 2006 Results Mark Scheme Edexcel GCE Chemistry 6246/02

2

Section A 1 (a) (i) pink to colourless NOT clear pink is decolourised ALLOW pale red instead of pink NOT pink goes on its own NOT purple NOT Red NOT any of these in combination with pink NOT Magenta NOT Cerise (1 mark) (ii) 8 less than 11 - Any number or range within this range (1 mark) (b) IGNE SF in (b) (i) initial no. moles NaOH = 1.00 x 25.0 = 0.0250 (1) 1000 no. moles HCl used = 0.100 x 8.80 = 8.80x10-4 (1) 1000 no. moles Na OH left in 25 cm 3 = 8.80x10-4 indication of 1:1 ratio (1) no. moles NaOH left in 250 cm 3 = 8.80x10-3 (1) no. moles NaOH used = 0.0250 8.80x10-3 = 0.0162 (1) 0.0162 with some working involving titre (5) Units not required BUT incorrect units e.g. mol dm -3 loses the 5 th mark (5 marks) (ii) no. moles hydrolysed = (b)(i) (1) = 8.1 x 10-3 2 EITHER mass of aspirin 8.1 x 10-3 x 180 = 1.458 g (1) % aspirin = 1.458 x 100 = 97.2 % (1) 1.50 Theoretical moles aspirin = 1.5 = 8.33 x 10-3 (1) 180 % aspirin = 8.1 x 10-3 x 100 = 97.2 % (1) 8.33 x 10-3 ALTERNATIVE METHOD Theoretical moles aspirin = 1.5 = 8.33 x 10-3 (1) Theoretical moles NaOH = 2 x 8.33 x 10-3 (1) = 0.01667 % aspirin = (b)(i) x 100 = 97.2 % (1) 0.01667 Mark consequentially but do not allow > 100 % 3

(c) (2 marks) Total for question:12 marks 4

Section B 2 (a) Molecular formula of D C 5 H 10 O 2 (2) with some correct working / deduction e.g. C 58.8 x 102 = 5 100 12 H 9.8 x 102 = 10 100 1 O 31.4 x 102 = 2 100 16 Use % to find empirical formula (1) then use or refer to molar mass to deduce molecular formula (1). E Is propan-2-ol (1) ACCEPT name or formula IF name and formula given, both must be correct Must contain CH 3 CH(OH)/be a secondary 2-ol/methyl secondary alcohol, as it gives iodoform ppt. (1) Do not allow if methyl ketone included G Is Iodoform/CHI 3 (1) stand alone mark IF name and formula given, both must be correct F Is the sodium salt of the acid/sodium ethanoate (produced by hydrolysis of the ester) (1) (so is sodium ethanoate) with justification for number of carbon atoms e.g. must contain 5-3=2 carbon atoms (1) ACCEPT name or formula ALLOW 1 max (out of 2) if ethanoic acid + reasoning for number of C atoms. 5

D is H O H H C C O C CH 3 1-methylethylethanoate CH 3 COOCH(CH 3 ) 2 CH 3 CO 2 CH(CH 3 ) 2 (1) H CH 3 ALLOW 2-propylethanoate D is consequential on their unambiguous E + F NOT just propylethanoate unless correct formula given Equation CH 3 COOCH(CH 3 ) 2 + NaOH CH 3 COONa + (CH 3 ) 2 CHOH (1) NOT molecular formula for D Consequential on their D ALLOW CH 3 COOC 3 H 7 for D and C 3 H 7 OH for E Candidates can identify D, E and F in the equation (9 marks) 6

(b) ALLOW correct names or formulae for reagents. If both given, both must be correct Condition mark only scores if correct or nearly correct reagents Step 1 CH 3 Cl / CH 3 COCl / any halogenoalkane/ any acylchloride (1) + (anhydrous) AlCl 3 / Al 2 Cl 6 / FeCl 3 / Fe 2 Cl 6 (1) ALLOW other halides except FeI 3 If RCl plus correct condition ALLOW condition mark Intermediate methylbenzene/c 6 H 5 CH 3 or other derived from their Freidel Crafts reagent (1) If step 1 uses HCOCl 4 max with -1 for each error. Step 2 KMnO 4 /potassium manganate(vii)/ MnO 4 (1) ALLOW potassium manganate, KMnO 4 NaOH/alkaline/OH (aq) and heat/reflux (1) NOT warm ACCEPT specified temperature provided it is 100 C Step 3 Acid/acidify / H + (aq)/named mineral acid (1) ALTERNATIVE Step 1 Br 2 any halogen (1) FeBr 3 etc (1) Intermediate bromobenzene (1) Step 2 Mg and dry ether (1) Step 3 (Solid) CO 2 (1) Step 4 Acid/water (1) ALLOW any correct synthesis with 1 for each error or omission made. If synthesis does not work, mark as follows: Start at beginning and mark until incorrect. Then start at end and mark until incorrect. Then award whichever route through gives the highest mark. 7 (6 marks)

(c) HNO 3 + H 2 SO 4 H 2 O + HSO 4 + NO + 2 - Can be shown in two stages HNO 3 + 2H 2 SO 4 H 3 O + + 2HSO 4 + NO + 2 (1) 1 st curved arrow from benzene ring of electrons towards N of NO 2 + ion (1) ALLOW the + anywhere on NO 2 Intermediate correctly drawn, including positive charge (1) Curved arrow from C-H bond back into benzene ring (1) IGNE if towards the + ALLOW HSO 4 but arrow must start on O ALLOW arrow from negative charge ALTERNATIVE 1 st curved arrow from double bond towards N of NO 2 + ion (1) ALLOW the + anywhere on NO 2 Intermediate correctly drawn, including positive charge (1) Curved arrow from C-H bond back into benzene ring (1) IGNE if towards the + (4 marks) Total for question 19 marks 8

3 (a) Q W C* Temperature 975 1225 K 700-950 o C (1) ALLOW any number or range within these values (Forward) reaction is exothermic so the highest yield should be at low temperature (1) reverse argument Rate is too slow at low temperature, so compromise (1) NOT just a compromise, it must be related to rate Catalyst Platinum (and rhodium) to give a fast rate at lower T Pt etc. speeds up conversion to NO and not combustion to N 2 (1) ALLOW Pt etc increases rate / lowers E a Pressure 2-10 atmospheres / 200 1000 kpa (1) stand alone ALLOW any number or range within these values (Small) increase/very little difference in number of (gas) molecules/moles on r.h.s. so low pressure (1) - stand alone 2-10 atmospheres / 200 1000 kpa (1) ALLOW any number or range within these values To push gases through (1) provided quoted or implied pressure (6 marks) (b) (i) C 4 H 4 O 6 2 + 5H 2 O 2 + 2H + 4CO 2 + 8H 2 O 1:5 ratio and no electrons in equation (1) Rest correct i.e. H + and H 2 O cancel (1) (2 marks) (ii) Rapid/fast effervescence ( rate has increased) (1) Colour change to green and back to pink/original colour ( alternative route/ not used up) (1) Do NOT allow if stated it does not take part in reaction NOT just back to pink NOT just pink colour returns They can change oxidation state easily/have variable oxidation states/can be illustrated (1) 9

(iii) Q W C* axes both labelled correctly (1) e.g. percentage/fraction/number of molecules (with energy E) for y-axis e.g. energy/kinetic energy for x-axis NOT speed shape of graph (1) starts at the origin (and rises steeply) peak skewed to left asymptote (if line crosses the x axis do not award this mark) All 3 needed Do not award the mark if two or more curves are drawn for different temperatures. Both activation energies shown well to the right of the peak (1) Comment on relationship of area under curve to number of particles with E E a e.g. more of molecules/collisions have energy greater than or equal to the activation energy/have enough energy to result in a reaction (1) Therefore a higher frequency of collisions result in reaction more (of the) collisions result in reaction more successful collisions per unit time more of the collisions are successful greater proportion of the collisions are successful (1) stand alone NOT just more successful collisions (5 marks) 10

(c) no. moles O 2 = 100 (1) 24000 = 4.17 x 10-3 no. moles H 2 O 2 = 2 x 4.17x10-3 (1) = 8.33x10-3 24000 cm 3 O 2 from 2 mol H 2 O 2 (1) therefore 100 cm 3 from 2 x 100 24000 = 8.33 x 10-3 mol (1) Mark consequentially on moles of O 2 volume H 2 O 2 = 8.33x10-3 x 1000 0.5 16.7 cm 3 (1) 8.33 x10-3 0.5 = 0.0167 dm 3 (1) Answer and unit required for third mark IGNE s.f. Mark consequentially on moles of H 2 O 2 Total for question: 19 marks 11

4 (a) (i) Fe 2 O 3 + 3CO 2Fe + 3CO 2 (1) H = (3 x 394) - (3 x 110) (-822) = -30 (kj mol -1 ) (2) Multiply by 3 twice (1) Correct answer with sign (1) ALLOW consequential calculation if wrong stoichiometry If O 2 given as product 1 max on consequential calculation (ii) (i) is more likely because the rate of a reaction between a solid and a gas will be faster than that between two solids ALLOW (i) is more likely because it is exothermic (and (ii) is endothermic) products in (i) are more thermodynamically stable relative to reactants than in (ii) Consequential on (a)(i) (1 mark) (b) (i) K p = ph 2 4 ph 2 O 4 (1) p s are essential NO [ ] = 1.6 4 = 3.16 / 3.2 and no units (1) 1.2 4 Consequential on K p expression provided no Fe or Fe 3 O 4 included (2 marks) (ii) Q W C* K P decreases (1) Because forward reaction release heat / exothermic reverse reaction absorbs heat / endothermic (1) Dependent on K p decreases Do not allow 2 ND mark if decrease is explained in terms of position moving to the left, UNLESS moving is a consequence of K p decreasing. (2 marks) (c) IGNE state symbols X: 2H 2 O + O 2 + 4e ( ) 4OH (1) ½ this multiples (1) Y: Fe Fe 2+ + 2e ( ) / Fe 2e ( ) Fe 2+ (1) IF X and Y not identified 1 (out of 2) iron(ii) hydroxide (1) ALLOW Fe(OH) 2 / [Fe(OH) 2 (H 2 O) 4 ] 12

(d) covalent bonds labelled (1) dative bonds labelled shown as arrows (1) ALLOW dot and cross diagram (2) Structure with Fe-Fe bond does not score first two marks Tetrahedral (around Fe) (1) ALLOW good 3-D diagram dependent on 4 covalent/dative bonds around Fe If Al 2 Cl 6 ALLOW max 2 (out of 3) Any mention of ionic 0 (out of 3) (e) (i) add (aqueous) sodium hydroxide / ammonia (1) ALLOW OH (aq) red/brown/foxy red/red-brown/rust ppt/solid (1) add (aqueous) potassium hexacyanoferrate(ii)/hexacyanoferrate(ii) ions (1) (Prussian) blue ppt/solid (1) ALLOW result for near miss spelling of reagent add (aqueous) potassium thiocyanate (1) blood red (solution) (1) NOT ppt (2 marks) (ii) Fe 3+ polarises the (OH bond in water) ligands (1) [Fe(H 2 O) 6 ] 3+ + H 2 O [Fe(OH)(H 2 O) 5 ] 2+ + H 3 O + in words e.g. deprotonation (of the ligand) by the (solvent) water (1) the H 3 O + / H + (aq) ions make the solution acidic (1) stand alone Total for question: 19 marks Total for paper: 50 marks 13